Question

A long wire carries a steady current. It is bent into circle of one turn and the magnetic field at the centre of the coil is $B$. It is then bent into a circular loop of $n$ turns. The magnetic field at the centre of the coil will be

• A. $nB$
• B. $n^{2}B$
• C. $2nB$
• D. $2n^{2}B$

Answer 1

The correct option is B $n^{2}B$

Let the length of the wire= $l$

Now, $2\pi r=l\Rightarrow r=\frac{l}{2\pi }$

At first magnetic field at the centre= $\frac{{\mu }_o}{4\pi }\frac{I}{r}2\pi$

$B=\frac{{\mu }_o}{2}\frac{I}{l}2\pi$

$B=\frac{{\mu }_{o}I\pi }{l}$

If the wire is bent into circular loop of $n$ turns.

Then, $2\pi r^{\prime }n=l$

$\Rightarrow r^{\prime }=\frac{l}{2\pi n}$

Magnetic field at its center= $B^{\prime }=\frac{{\mu }_o}{4\pi }\frac{I}{r^{\prime }}2\pi n$

$=\frac{{\mu }_o}{2}\frac{I}{l}2\pi n\times n$

$=\frac{{\mu }_{o}I\pi }{l}{n}^2$

$\Rightarrow B^{\prime }=n^{2}B$ .